Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 1e5 + 10;
long long n, m, q, u, v, c, a, lca, l = 20, h[MAXN], par[20 + 10][MAXN];
vector<long long> adj[MAXN], id[MAXN], ans[20 + 10][MAXN], vctr;
vector<long long> merge(vector<long long> vec1, vector<long long> vec2) {
  long long ptr = 0, ptr2 = 0;
  vector<long long> vec;
  while (ptr < (long long)vec1.size() || ptr2 < (long long)vec2.size())
    if (ptr == (long long)vec1.size())
      vec.push_back(vec2[ptr2++]);
    else if (ptr2 == (long long)vec2.size())
      vec.push_back(vec1[ptr++]);
    else {
      if (vec1[ptr] < vec2[ptr2])
        vec.push_back(vec1[ptr++]);
      else if (vec1[ptr] > vec2[ptr2])
        vec.push_back(vec2[ptr2++]);
      else
        vec.push_back(vec1[ptr++]), ptr2++;
    }
  while ((long long)vec.size() > 10) vec.pop_back();
  return vec;
}
void dfs(long long u, long long p) {
  for (auto v : adj[u])
    if (v != p) {
      h[v] = h[u] + 1, par[0][v] = u, ans[0][v] = merge(id[v], id[u]);
      for (long long i = 1; i < l; i++)
        par[i][v] = par[i - 1][par[i - 1][v]],
        ans[i][v] = merge(ans[i - 1][v], ans[i - 1][par[i - 1][v]]);
      dfs(v, u);
    }
  return;
}
long long Find_LCA(long long u, long long x) {
  for (long long i = l - 1; i >= 0; i--)
    if (x & (1 << i)) u = par[i][u];
  return u;
}
vector<long long> get_ans(long long u, long long x) {
  vector<long long> vec;
  for (long long i = l - 1; i >= 0; i--)
    if (x & (1 << i)) vec = merge(vec, ans[i][u]), u = par[i][u];
  return vec;
}
long long LCA(long long u, long long v) {
  if (h[v] < h[u]) swap(v, u);
  long long x = h[v] - h[u];
  v = Find_LCA(v, x);
  if (u == v) return v;
  for (long long i = l - 1; i >= 0; i--)
    if (par[i][v] != par[i][u]) u = par[i][u], v = par[i][v];
  return par[0][u];
}
void input() {
  cin >> n >> m >> q;
  for (long long i = 1; i < n; i++)
    cin >> u >> v, u--, v--, adj[u].push_back(v), adj[v].push_back(u);
  for (long long i = 0; i < m; i++) cin >> c, c--, id[c].push_back(i + 1);
  return;
}
int32_t main() {
  input();
  for (long long i = 0; i < n; i++) {
    sort(id[i].begin(), id[i].end());
    while ((long long)id[i].size() > 10) id[i].pop_back();
  }
  dfs(0, -1);
  while (q--) {
    cin >> u >> v >> a, u--, v--;
    if (u == v) {
      cout << min((long long)id[u].size(), a) << " ";
      for (long long i = 0; i < min((long long)id[u].size(), a); i++)
        cout << id[u][i] << " ";
      cout << '\n';
      continue;
    }
    lca = LCA(u, v),
    vctr = merge(get_ans(u, h[u] - h[lca]), get_ans(v, h[v] - h[lca])),
    cout << min((long long)vctr.size(), a) << " ";
    for (long long i = 0; i < min((long long)vctr.size(), a); i++)
      cout << vctr[i] << " ";
    cout << endl;
  }
  return 0;
}